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Difference between revisions of "2023 AMC 10B Problems/Problem 14"

(Solution 4)
m (→‎Solution 2)
 
(22 intermediate revisions by 13 users not shown)
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== Solution 1 ==
 
== Solution 1 ==
  
−
Clearly, <math>m=0,n=0</math> is 1 solution. However there are definitely more, so we apply [https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Expression] to get this:
+
Let's use 10th grade math to solve this. After all, it is called the AMC 10 for a reason!
 +
 
 +
We have
 +
<cmath>
 +
m^2 + mn + n^2 = m^2n^2.
 +
</cmath>
 +
We subtract \( mn \) on both sides to get
 +
 
 +
\( m^2 + n^2 = m^2n^2 - mn \).
 +
 
 +
Fun Fact! You can write \( m^2 + n^2 \) as \( (m+n)^2 - 2mn \)! Let's use this!
 +
 
 +
We convert the Left Hand Side into \( (m+n)^2 - 2mn \) to get
 +
\( (m+n)^2 - 2mn = m^2n^2 - mn \).
 +
 
 +
Adding by \( 2mn \) gives us \( (m+n)^2 = m^2n^2 + mn \).
 +
 
 +
We aren't done yet though! \( m^2n^2 + mn \) can be simplified into \( mn(mn+1) \), giving us \( (m+n)^2 = mn(mn+1) \).
 +
 
 +
Okay so now we're done, but, Pinotation, this doesn't do anything! Well, now we can use the Zero Product Property!
 +
 
 +
How? I'll show you!
 +
 
 +
We subtract \( mn(mn+1) \) on both sides of the equation to get \( (m+n)^2 - mn(mn+1) = 0 \). Now we do a bit of casework.
 +
 
 +
Notice how \( (m+n)^2 - mn(mn+1) = 0 \) is just \( (m+n)(m+n) - mn(mn+1) = 0 \). So, either \( m+n = 0 \) and \( mn = 0 \), or \( m +n = 0 \) and \( mn+1 = 0 \). Let's look at it through both cases.
 +
 
 +
Case 1: \( m+n = 0 \) and \( mn = 0 \). If \( m+n = 0 \) and \( mn = 0 \), then that must mean that either \( m = 0 \) or \( n = 0 \), and if we substitute either \( m=0 \) or \( n=0 \) in, we still get either \( m=0 \) or \( n=0 \), so therefore we have 1 ordered pair, \( (0,0) \).
 +
 
 +
Case 2: \( m +n = 0 \) and \( mn+1 = 0 \). \( mn+1 =0 \) means that \( mn=-1 \). So, for this to be possible, either \( m = -1 \) and \( n=1 \), or \( m=1 \) and \( n=-1 \). Let's check for contradictions quickly. We see that \( m + n = 0 \), and \( -1 + 1 = 0 \) and \( 1 - 1 =0 \), so we know the ordered pairs \( (-1,1) \) and \( (1,-1) \) both work.
 +
 
 +
We have a total of <math>\boxed{\textbf{(C) 3}}.</math> ordered pairs. \( (-1,1) \), \( (0,0) \), and \( (1,-1) \).
 +
 
 +
~Pinotation
 +
 
 +
==Solution 2==
 +
 
 +
Clearly, <math>m=0,n=0</math> is one of the solutions. However, we can be quite sure that there are more, so we apply [https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Trick] to get the following:  
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 12: Line 49:
 
m^2+mn+n^2 +mn &= m^2n^2 +mn\\
 
m^2+mn+n^2 +mn &= m^2n^2 +mn\\
 
(m+n)^2 &= m^2n^2 +mn\\
 
(m+n)^2 &= m^2n^2 +mn\\
−
(m+n)^2 &= mn(mn+1)\\
+
(m+n)^2 &= mn(mn+1).\\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
−
This basically say that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square which is practically impossible except <math>mn=0</math> or <math>mn+1=0</math>.
+
Essentially, this says that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square. This is impossible except for when <math>mn=0</math> or <math>mn+1=0</math>.  
 
<math>mn=0</math> gives <math>(0,0)</math>.
 
<math>mn=0</math> gives <math>(0,0)</math>.
−
<math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>.
+
<math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>. Answer: <math>\boxed{\textbf{(C) 3}}.</math>
  
−
~Technodoggo ~minor edits by lucaswujc
+
~Technodoggo ~minor edits by lucaswujc ~minor edits by luke22
  
−
==Solution 2 ==
+
==Solution 3 ==
  
 
Case 1: <math>mn = 0</math>.
 
Case 1: <math>mn = 0</math>.
Line 41: Line 78:
  
 
Modulo <math>u</math>, we have <math>v^2 \equiv 0 \pmod{u}</math>.
 
Modulo <math>u</math>, we have <math>v^2 \equiv 0 \pmod{u}</math>.
−
Because <math>\left( u, v \right) = 1</math>, we must have <math>|u| = |v| = 1</math>.
+
Because <math>{\rm gcd} \left( u, v \right) = 1</math>., we must have <math>|u| = |v| = 1</math>.
 
Plugging this into the above equation, we get <math>2 + uv = k^2</math>.
 
Plugging this into the above equation, we get <math>2 + uv = k^2</math>.
 
Thus, we must have <math>uv = -1</math> and <math>k = 1</math>.
 
Thus, we must have <math>uv = -1</math> and <math>k = 1</math>.
Line 51: Line 88:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
~ sravya_m18
  
−
==Solution 3 (Discriminant)==
+
==Solution 4 (Discriminant)==
−
We can move all terms to one side and wrote the equation as a quadratic in terms of <math>n</math> to get <cmath>(1-m^2)n^2+(m)n+(m^2)=0.</cmath> The discriminant of this quadratic is <cmath>\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).</cmath> For <math>n</math> to be an integer, we must have <math>m^2(4m^2-3)</math> be a perfect square. Thus, either <math>4m^2-3</math> is a perfect square or <math>m^2 = 0</math> and <math>m = 0</math>. The first case gives <math>m=-1,1</math>, which result in the equations <math>-n+1=0</math> and <math>n-1=0</math>, for a total of two pairs: <math>(-1,1)</math> and <math>(1,-1)</math>. The second case gives the equation <math>n^2=0</math>, so it's only pair is <math>(0,0)</math>. In total, the total number of solutions is <math>\boxed{\textbf{(C) 3}}</math>.
+
We can move all terms to one side and write the equation as a quadratic in terms of <math>n</math> to get <cmath>(1-m^2)n^2+(m)n+(m^2)=0.</cmath> The discriminant of this quadratic is <cmath>\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).</cmath> For <math>n</math> to be an integer, we must have <math>m^2(4m^2-3)</math> be a perfect square. Thus, either <math>(2m)^2-3</math> is a perfect square or <math>m^2 = 0</math> and <math>m = 0</math>. The first case gives <math>m=-1,1</math> (larger squares are separated by more than 3), which result in the equations <math>-n+1=0</math> and <math>n-1=0</math>, for a total of two pairs: <math>(-1,1)</math> and <math>(1,-1)</math>. The second case gives the equation <math>n^2=0</math>, so it's only pair is <math>(0,0)</math>. In total, the total number of solutions is <math>\boxed{\textbf{(C) 3}}</math>.
  
 
~A_MatheMagician
 
~A_MatheMagician
  
−
==Solution 4 (Nice Substitution)==
+
==Solution 5 (Nice Substitution)==
−
Let <math>x=m+n, y=xn</math> then <cmath>x^2-y=y^2</cmath> Completing the square then gives <cmath>4x^2+1=(2y+1)^2</cmath> Since the RHS is a square, clearly the only solutions are <math>x=0,y=0</math> and <math>x=0,y=-1</math>. The first gives <math>(0,0)</math> while the second gives <math>(-1,1)</math> and <math>(1,-1)</math> by solving it as a quadratic with roots <math>m</math> and <math>n</math>. Thus there are <math>\boxed{\textbf{(C) 3}}</math> solutions.
+
Let <math>x=m+n, y=mn</math> then  
 +
<cmath>x^2-y=y^2</cmath>
 +
 
 +
Completing the square in <math>y</math> and multiplying by 4 then gives
 +
<cmath>4x^2+1=(2y+1)^2</cmath>  
 +
 
 +
Since the RHS is a square, clearly the only solutions are <math>x=0,y=0</math> and <math>x=0,y=-1</math>.
 +
 
 +
The first gives <math>(0,0)</math>.
 +
 
 +
The second gives <math>(-1,1)</math> and <math>(1,-1)</math> by solving it as a quadratic with roots <math>m</math> and <math>n</math>.  
 +
 
 +
Thus there are <math>\boxed{\textbf{(C) 3}}</math> solutions.
  
 
~ Grolarbear
 
~ Grolarbear
 +
 +
==Solution 6 (Alternative Method for Manipulation)==
 +
 +
<math>m^2 + mn + n^2 = m^2n^2</math>
 +
 +
<math>mn = m^2n^2 - m^2 - n^2</math>
 +
 +
<math>mn + 1 = m^2(n^2 - 1) - 1(n^2 - 1)</math>
 +
 +
<math>mn + 1 = (m + 1)(m - 1)(n + 1)(n - 1)</math>
 +
 +
Notice that the right side can be zero or one.
 +
If the right side is zero, m and n can be <math>(-1,1)</math> and <math>(1,-1)</math>.
 +
If the right side is one, m and n can be <math>(0,0)</math>.
 +
There are <math>\boxed{\textbf{(C) 3}}</math> solutions.
 +
 +
~unhappyfarmer
 +
 +
==Solution 7 (Obtaining ranges)==
 +
 +
Set <math>m\leq n</math>. Then, we can say that
 +
 +
<cmath>3n^2\geq m^2n^2</cmath>
 +
 +
<cmath>3\geq m^2</cmath>
 +
 +
Or <math>-\sqrt{3} \leq m \leq \sqrt{3}</math>, and since we are dealing with integers, <math>m=-1</math>, <math>0</math> or <math>1</math>. Testing these numbers, we get <math>n=1</math>, <math>n=0</math> and <math>n=-1</math> respectively. Although the solution <math>(1,-1)</math> is a solution in the end, our initial condition for this case was <math>m\leq n</math>. For better rigour, we can just consider the other case <math>m>n</math> to validate solution <math>(1,-1)</math>.
 +
 +
-lisztepos
 +
 +
==Solution 8 (Inequality)==
 +
 +
If <math>mn = 0</math>, then <math>m^2 + 0 + n^2 = 0</math>, <math>(m, n) = (0, 0)</math>.
 +
 +
If <math>mn \neq 0</math>, then
 +
 +
<cmath>\frac{m^2 + mn + n^2}{m^2n^2} = 1</cmath>
 +
 +
<cmath>\frac{1}{m^2} + \frac{1}{mn} + \frac{1}{n^2} = 1</cmath>
 +
 +
<cmath>\frac{1}{m^2} + \frac{2}{|mn|} + \frac{1}{n^2} > 1</cmath>
 +
 +
<cmath>\left(\frac{1}{|m|} + \frac{1}{|n|}\right)^2 > 1</cmath>
 +
 +
<cmath>\frac{1}{|m|} + \frac{1}{|n|} > 1</cmath>
 +
 +
Obviously, at least one of <math>|m|, |n|</math> is 1. If <math>m = 1</math>, <math>1 + n + n^2 = n^2 \Rightarrow n = -1</math>. If <math>m = -1</math>, <math>1 - n + n^2 = n^2 \Rightarrow n = 1</math>. We omit the discussion of <math>n = \pm 1</math>.
 +
 +
Finally, we get <math>(m, n) = (0, 0), (1, -1), (-1, 1)</math>, there are <math>\boxed{\textbf{(C) 3}}</math> solutions.
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Reda_mandymath reda_mandymath]
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==
Line 70: Line 171:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
==Video Solution by Interstigation==
  
 +
https://youtu.be/Vq7kevsWlHk
 +
 +
~Interstigation
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2023|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2023|ab=B|num-b=13|num-a=15}}
 +
 +
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:00, 13 September 2025

Problem

How many ordered pairs of integers $(m, n)$ satisfy the equation $m^2+mn+n^2 = m^2n^2$?

$\textbf{(A) }7\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }5$

Solution 1

Let's use 10th grade math to solve this. After all, it is called the AMC 10 for a reason!

We have \[m^2 + mn + n^2 = m^2n^2.\] We subtract \( mn \) on both sides to get

\( m^2 + n^2 = m^2n^2 - mn \).

Fun Fact! You can write \( m^2 + n^2 \) as \( (m+n)^2 - 2mn \)! Let's use this!

We convert the Left Hand Side into \( (m+n)^2 - 2mn \) to get \( (m+n)^2 - 2mn = m^2n^2 - mn \).

Adding by \( 2mn \) gives us \( (m+n)^2 = m^2n^2 + mn \).

We aren't done yet though! \( m^2n^2 + mn \) can be simplified into \( mn(mn+1) \), giving us \( (m+n)^2 = mn(mn+1) \).

Okay so now we're done, but, Pinotation, this doesn't do anything! Well, now we can use the Zero Product Property!

How? I'll show you!

We subtract \( mn(mn+1) \) on both sides of the equation to get \( (m+n)^2 - mn(mn+1) = 0 \). Now we do a bit of casework.

Notice how \( (m+n)^2 - mn(mn+1) = 0 \) is just \( (m+n)(m+n) - mn(mn+1) = 0 \). So, either \( m+n = 0 \) and \( mn = 0 \), or \( m +n = 0 \) and \( mn+1 = 0 \). Let's look at it through both cases.

Case 1: \( m+n = 0 \) and \( mn = 0 \). If \( m+n = 0 \) and \( mn = 0 \), then that must mean that either \( m = 0 \) or \( n = 0 \), and if we substitute either \( m=0 \) or \( n=0 \) in, we still get either \( m=0 \) or \( n=0 \), so therefore we have 1 ordered pair, \( (0,0) \).

Case 2: \( m +n = 0 \) and \( mn+1 = 0 \). \( mn+1 =0 \) means that \( mn=-1 \). So, for this to be possible, either \( m = -1 \) and \( n=1 \), or \( m=1 \) and \( n=-1 \). Let's check for contradictions quickly. We see that \( m + n = 0 \), and \( -1 + 1 = 0 \) and \( 1 - 1 =0 \), so we know the ordered pairs \( (-1,1) \) and \( (1,-1) \) both work.

We have a total of $\boxed{\textbf{(C) 3}}.$ ordered pairs. \( (-1,1) \), \( (0,0) \), and \( (1,-1) \).

~Pinotation

Solution 2

Clearly, $m=0,n=0$ is one of the solutions. However, we can be quite sure that there are more, so we apply Simon's Favorite Factoring Trick to get the following:

\begin{align*} m^2+mn+n^2 &= m^2n^2\\ m^2+mn+n^2 +mn &= m^2n^2 +mn\\ (m+n)^2 &= m^2n^2 +mn\\ (m+n)^2 &= mn(mn+1).\\ \end{align*}

Essentially, this says that the product of two consecutive numbers $mn,mn+1$ must be a perfect square. This is impossible except for when $mn=0$ or $mn+1=0$. $mn=0$ gives $(0,0)$. $mn=-1$ gives $(1,-1), (-1,1)$. Answer: $\boxed{\textbf{(C) 3}}.$

~Technodoggo ~minor edits by lucaswujc ~minor edits by luke22

Solution 3

Case 1: $mn = 0$.

In this case, $m = n = 0$.

Case 2: $mn \neq 0$.

Denote $k = {\rm gcd} \left( m, n \right)$. Denote $m = k u$ and $n = k v$. Thus, ${\rm gcd} \left( u, v \right) = 1$.

Thus, the equation given in this problem can be written as \[ u^2 + uv + v^2 = k^2 u^2 v^2 . \]

Modulo $u$, we have $v^2 \equiv 0 \pmod{u}$. Because ${\rm gcd} \left( u, v \right) = 1$., we must have $|u| = |v| = 1$. Plugging this into the above equation, we get $2 + uv = k^2$. Thus, we must have $uv = -1$ and $k = 1$.

Thus, there are two solutions in this case: $\left( m , n \right) = \left( 1, -1 \right)$ and $\left( m , n \right) = \left( -1, 1 \right)$.

Putting all cases together, the total number of solutions is $\boxed{\textbf{(C) 3}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) ~ sravya_m18

Solution 4 (Discriminant)

We can move all terms to one side and write the equation as a quadratic in terms of $n$ to get \[(1-m^2)n^2+(m)n+(m^2)=0.\] The discriminant of this quadratic is \[\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).\] For $n$ to be an integer, we must have $m^2(4m^2-3)$ be a perfect square. Thus, either $(2m)^2-3$ is a perfect square or $m^2 = 0$ and $m = 0$. The first case gives $m=-1,1$ (larger squares are separated by more than 3), which result in the equations $-n+1=0$ and $n-1=0$, for a total of two pairs: $(-1,1)$ and $(1,-1)$. The second case gives the equation $n^2=0$, so it's only pair is $(0,0)$. In total, the total number of solutions is $\boxed{\textbf{(C) 3}}$.

~A_MatheMagician

Solution 5 (Nice Substitution)

Let $x=m+n, y=mn$ then \[x^2-y=y^2\]

Completing the square in $y$ and multiplying by 4 then gives \[4x^2+1=(2y+1)^2\]

Since the RHS is a square, clearly the only solutions are $x=0,y=0$ and $x=0,y=-1$.

The first gives $(0,0)$.

The second gives $(-1,1)$ and $(1,-1)$ by solving it as a quadratic with roots $m$ and $n$.

Thus there are $\boxed{\textbf{(C) 3}}$ solutions.

~ Grolarbear

Solution 6 (Alternative Method for Manipulation)

$m^2 + mn + n^2 = m^2n^2$

$mn = m^2n^2 - m^2 - n^2$

$mn + 1 = m^2(n^2 - 1) - 1(n^2 - 1)$

$mn + 1 = (m + 1)(m - 1)(n + 1)(n - 1)$

Notice that the right side can be zero or one. If the right side is zero, m and n can be $(-1,1)$ and $(1,-1)$. If the right side is one, m and n can be $(0,0)$. There are $\boxed{\textbf{(C) 3}}$ solutions.

~unhappyfarmer

Solution 7 (Obtaining ranges)

Set $m\leq n$. Then, we can say that

\[3n^2\geq m^2n^2\]

\[3\geq m^2\]

Or $-\sqrt{3} \leq m \leq \sqrt{3}$, and since we are dealing with integers, $m=-1$, $0$ or $1$. Testing these numbers, we get $n=1$, $n=0$ and $n=-1$ respectively. Although the solution $(1,-1)$ is a solution in the end, our initial condition for this case was $m\leq n$. For better rigour, we can just consider the other case $m>n$ to validate solution $(1,-1)$.

-lisztepos

Solution 8 (Inequality)

If $mn = 0$, then $m^2 + 0 + n^2 = 0$, $(m, n) = (0, 0)$.

If $mn \neq 0$, then

\[\frac{m^2 + mn + n^2}{m^2n^2} = 1\]

\[\frac{1}{m^2} + \frac{1}{mn} + \frac{1}{n^2} = 1\]

\[\frac{1}{m^2} + \frac{2}{|mn|} + \frac{1}{n^2} > 1\]

\[\left(\frac{1}{|m|} + \frac{1}{|n|}\right)^2 > 1\]

\[\frac{1}{|m|} + \frac{1}{|n|} > 1\]

Obviously, at least one of $|m|, |n|$ is 1. If $m = 1$, $1 + n + n^2 = n^2 \Rightarrow n = -1$. If $m = -1$, $1 - n + n^2 = n^2 \Rightarrow n = 1$. We omit the discussion of $n = \pm 1$.

Finally, we get $(m, n) = (0, 0), (1, -1), (-1, 1)$, there are $\boxed{\textbf{(C) 3}}$ solutions.

~reda_mandymath

Video Solution by OmegaLearn

https://youtu.be/5a5caco_YTo

Video Solution

https://youtu.be/Dh1lDI1fHrw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/Vq7kevsWlHk

~Interstigation

See also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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